Math for Assembly

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Given the following segment:offset addresses, find the physical address.

Formula:\quad physical\ address = (segment\times 16) + offset

(a) 02C1:074F

\begin{aligned} 02C10& \\ +\quad 074F& \\ \overline{0335F}& \end{aligned}

(b) 15C5:4CCF

\begin{aligned} \tt 15C50& \\ +\quad \tt 4CCF& \\ \overline{\tt 1A91F}& \end{aligned}

Given the following physical addresses and offset values, calculate the segment value.

Rearrange \\ Formula:\quad (physical\ address -\ offset) /16 = segment

(a) Physical Address C4AD6, Offset 8AB6

\begin{aligned} \tt C4AD6& \\ -\quad \tt 8AB6& \\ \overline{\tt BC020}& \\ \text{ Divided by E means physical address is BC02} \end{aligned}

(b) Physical Address 1B0DC, Offset 007C

\begin{aligned} \tt 1B0DC& \\ -\quad \tt 007C& \\ \overline{\tt 1B060}& \\ \text{ Divided by E means physical address is 1B06} \end{aligned}

Convert the following decimal reals to IEEE Single Precision binary.

(a) -9.8
The number is negative number therefore signed with $1$
$9$ in binary is $1001$

\begin{aligned} &0.8 \times 2 = 1.6 = 1\\ &0.6 \times 2 = 1.2 = 1\\ &0.2 \times 2 = 0.4 = 0\\ &0.4 \times 2 = 0.8 = 0\\ &0.8 \times 2 = 1.6 = 1\\ &\\ &0.8 = 0.1100110011001100... \end{aligned}

In normalized form $1.00101 \times 2^3$
Exponent of 3 needs to add 127 to get 130 which in binary is $1000 0010$
Thusly $\textcolor{khaki}{\text{-9.8 in floating point binary is \boxed{\text{1 10000010 0010 1000 0000 0000 0000 000 }}}}$

(b) 12.32
The number is positive number therefore signed with $0$
$12$ In binary the number is $1100$

\begin{align} &0.32 \times 2 = 0.64 = 0\\ &0.64 \times 2 = 1.28 = 1\\ &0.28 \times 2 = 0.56 = 0\\ &0.56 \times 2 = 1.12 = 1\\ &0.12 \times 2 = 0.24 = 0\\ &0.24 \times 2 = 0.48 = 0\\ &0.48 \times 2 = 0.96 = 0\\ &0.96 \times 2 = 1.92 = 1\\ &0.92 \times 2 = 1.84 = 1\\ &0.84 \times 2 = 1.68 = 1\\ &0.68 \times 2 = 1.36 = 1\\ &0.36 \times 2 = 0.72 = 0\\ &0.72 \times 2 = 1.44 = 1\\ &0.44 \times 2 = 0.88 = 0\\ &0.88 \times 2 = 1.76 = 1\\ &0.76 \times 2 = 1.52 = 1\\ &0.52 \times 2 = 1.04 = 1\\ &0.04 \times 2 = 0.08 = 0\\ &0.08 \times 2 = 0.16 = 0\\ &0.16 \times 2 = 0.32 = 0\\ &0.32 \times 2 = 0.64 = 0\\ &\\ &0.32 = 0.010100011110101110000... \end{align}

Exponent of 3 needs to add 127 to get 130 which in binary is $1000 0010$
In normalized form $1. 1000 1010 0011 11010 1110 000 \times 2^3$
Thusly $\textcolor{khaki}{\text{12.32 in floating point binary is \boxed{\text{0 10000010 1000 1010 0011 11010 1110 000 }}}}$

Convert the following IEEE Single Precision binary numbers to decimal real values.
(a) 1 01111100 1000 0001 1000 1011 1011 100

The number is signed with $1$ therefore it is a negative number

Exponent is $01111100$ which is

$\displaystyle \begin{aligned} &\quad 2^6 &+\quad 2^5 &+\quad 2^4 &+\quad 2^3 &+\quad 2^2 \\ &\quad 64 &+\quad 32 &+\quad 16 &+\quad 8 &+\quad 4 \quad= 124 \end{aligned}$

$\displaystyle$

this is biased unbiased is $124 - 127 = -3$

so $\text{1.1000 0001 1000 1011 1011 100}\times 2^{-3}\\ \text{0.0011 0000 0011 0001 0111 0111 00}$ Denormalized binary form

2^{-4} + 2^{-9} + 2^{-8} + 2^{-16} + 2^{-18} + 2^{-19} + 2^{-20} + 2^{-22} + 2^{-23} + 2^{-24}\\ \textcolor{khaki}{\text{1 01111100 1000 0001 1000 1011 1011 100}}\\ \textcolor{khaki}{\text{in Decimal Form is}}\\ \textcolor{khaki}{\boxed{-0.1882547736168}}

(b) 0 10001100 0000 1011 1011 1011 1011 101

The number is signed with $0$ therefore it is a positive number

Exponent is $10001100$ which is

$\displaystyle \begin{aligned} &\quad 2^7 &+\quad 2^3&+\quad 2^2 \\ &\quad 128 &+\quad 8\ \ &+\quad 4 \quad= 140 \end{aligned}$

$\displaystyle$

this is biased unbiased is $140 - 127 = 13$

so $\text{1.0000 1011 1011 1011 1011 101}\times 2^{13}\\ \text{10 0001 0111 0111 . 0111 0111 01}$ Denormalized binary form

2^{13} + 2^8 + 2^6 + 2^5 + 2^4 + 2^2 + 2^1 + 2^0 + 2^{-2} + 2^{-3} + 2^{-4} + 2^{-6} + 2^{-7} + 2^{-8} + 2^{-10}\\ \textcolor{khaki}{\text{0 10001100 0000 1011 1011 1011 1011 101}}\\ \textcolor{khaki}{\text{in Decimal Form is}}\\ \textcolor{khaki}{\boxed{8567.2783203125}}